Say you have a column $C_i$ with negative sum, and you change the sign of all elements in that column. This must increase the total sum of all the elements in the grid (hereby just called the "total sum"), since all the other columns are unaffected, and the sum of all the column sums is the same as the total sum. The same thing goes for rows, of course.
There are only finitely many different total sums that can be made with the numbers on the board by changing the signs, even if you could freely change the signs of single elements. As long as there is a row or column with negative sum, you can increase the total sum, and increasing the total sum can be done only finitely many times. That means that the algorithm "Find a row or column with negative sum and flip the signs in it, then repeat" must end after a finite number of repeats.
Key Observation
Parity of the sum of the numbers written on the board stays unchanged. Because, at each step you are turning $x,y$ into $|x-y|$ and $x+y\equiv x-y\equiv y-x\pmod2$ .
Necessary condition
In order to have the last number to be $0$, you need the sum of the numbers written in the beginning to be even. So, we want
$$1+2+\ldots+n=\frac{n(n+1)}2$$
to be even. Notice, this happens iff $n\equiv 0,3\pmod4$.
To see this condition is sufficient, try to play with the numbers to get $0$ at the end. You said you did it for $4$. For $3$,
$$1,(2,3)\mapsto1,(1)\mapsto 0$$
works.
Suppose you can get $0$ for $n$. Then, you can get $0$ for $n+4$ by repeating the same steps for the first $n$ numbers to get $$0,n+1,n+2,n+3,n+4$$ and then making $$n+4,n+2\mapsto 2\qquad n+3,n+1\mapsto 2\qquad 2,2\mapsto 0\qquad 0,0\mapsto 0$$
So, by induction, you can do it for any $n\equiv 0,3\pmod4$
Best Answer
Any integer and it's negative is of same parity. $$a \equiv -a \pmod 2$$
We see $$\pm a_1 \pm a_2 \pm a_3 \ldots \pm a_n \equiv a_1 + a_2 + a_3 + \ldots a_n \pmod 2$$
Any combination of addition, subtraction results in same parity as all additions.